Question

The number of ways in which ten candidates $A1,A2,....A10$ can be ranked such that $A1$ is always above $A10$ is?
1.$5!$
2.$2\left( 5! \right)$
3.$10!$
4.$\dfrac{10!}{2}$

Answer 1

In order to find the number of ways in $A1$ is always above $A10$, firstly we will be calculating all the number of ways in which the ten candidates $A1,A2,....A10$ can be arranged and then we will be calculate the number of ways the required arrangement can be done and that would be our required answer.

Complete step by step answer:
Now let us briefly discuss the combinations. Combinations are selection of items from a group of items when the order of the selection is not considered. Combination simply deals with the selection. The notation of the combination is $^{n}{{C}_{r}}$. The formula for finding the number of combinations is $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. There are two basic principles of combinations. They are: Fundamental Principle of Counting and Addition Principle.
Now let us find out the number of ways in which ten candidates $A1,A2,....A10$ can be ranked such that $A1$ is always above $A10$.
Firstly, let us find out the number of ways in which the ten candidates can be arranged.
$\Rightarrow A1,A2,....A10=10!$
The number of ways in which $A1$ is always above $A10$ is $\dfrac{10!}{2}$
Because in the other half of the combination, $A10$ is always above $A1$.

So, the correct answer is “Option 4”.

Note: In the above case, only half of the combination is considered because the other half wouldn’t satisfy the given condition. We must always calculate according to the condition required. If the order is not specified, then we can apply the combination concept. Not applying the proper concept is the commonly committed error.